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Homework Help: Water droplet coalescence - prove S < S1 + S2

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    There are two water droplets with the radii R1 and R2, volumes V1 and V2, and surface areas S1 and S2. Assume that R1 is not equal to R2.

    Upon coalescence, two droplets form a "joint" droplet with the volume V=V1+V2.

    Prove analytically that the surface area of the "joint" droplet S < S1 + S2.

    2. Relevant equations
    surface area of a sphere, volume of a sphere, and radius of a sphere.

    3. The attempt at a solution

    So far I have, but not sure how to factor this out and turn into the inequality proof:

    S < S1 + S2
    4piR^2 < 4piR1^2 + 4piR2^2

    V = V1 + V2
    4/3piR^3 = 4/3piR1^3 + 4/3piR2^3
  2. jcsd
  3. Mar 6, 2016 #2
    is there one more condition related to surface tension?
  4. Mar 6, 2016 #3
    Assumed that pressure is equivalent
  5. Mar 6, 2016 #4
    Is it wise assumption?

    i do not think so - as any water droplet is having a pressure related to surface tension and radius r . you must use it
  6. Mar 6, 2016 #5
    What has the surface tension got to do with it if the shape is a sphere and the final volume is the sum of the two initial volumes?
  7. Mar 6, 2016 #6
    This is really a math problem, and I'm moving it to the Introductory Math Forum.

  8. Mar 6, 2016 #7
    actually i remember that droplets or bubbles are in equilibrium because of an excess pressure inside it which is due to the surface stretched to make the shape - its related to (2.surface tension/radius)so i thought that can give a relationship to get out of the jam'
  9. Mar 6, 2016 #8
    I don't see where there is a "jam." It's strictly a mathematics problem.
  10. Mar 6, 2016 #9
    Just substitute R from the volume equation for R in the inequality, and determine if it is satisfied.
  11. Mar 6, 2016 #10
    well ,i thought its related to idea of surface tension- it might have helped
  12. Mar 6, 2016 #11


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    But the problem does not have a thing about surface tension in it, so it doesn't help. It just adds confusion.
  13. Mar 6, 2016 #12


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    Bear in mind that the second is what you are given and the first is what you are trying to show, but we can cross that bridge later.
    Can you simplify those two equations?
    Can you combine them to eliminate a variable?
  14. Mar 6, 2016 #13
    actually liquid drops are made/exist in spherical shape due to surface tension property- in general one can not have water drops of different radius possible . they may flow out - it may not form or stay in spherical shape- for a drop to exist the pressure inside must be greater than outside pressure by an amount 2.T/r-
    but the present problem may be a purely 'mathematical' one -as @Chestermiller opined.
  15. Mar 6, 2016 #14


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    Your determination to make surface tension relevant somehow reminds me of a classic BBCTV sitcom.
    Basil Fawlty: What are you a doctor of?
    Hotel Guest: Paediatrics
    Fawlty: Feet?
    Guest: Children
    Fawlty: Well, children have feet, don't they?
  16. Mar 7, 2016 #15


    Staff: Mentor

    That's my opinion, as well. The droplets assume a spherical shape because of surface tension, but that's all you need to say about surface tension. The problem is strictly a matter of geometry, comparing the volumes and surface areas of droplets before and after they coalesce.

    Bringing up surface tension (several times) is really nothing more than a distraction in this problem.
  17. Mar 7, 2016 #16
    Yes, I believe this boils down to a math problem for me at this point. I understand the concepts but it's been a long time for me since I have been in school so I'm still brushing up on my math. I'll spend some time working on understanding solving the inequality and substitution. Thank you all for the input. If anyone can direct me to what I should brush up on the math part to solve this. I have my algebra/calc book out reading inequalities now.
  18. Mar 7, 2016 #17
    So I take the V equation and solve it for R, then plug that in for R into the S equation. Got it. Working on it now...
  19. Mar 7, 2016 #18
    So I have the following, and got a bit stuck.

    V = V1 + V2

    V = 4/3piR^3

    4/3piR^3 = 4/3piR1^3 + 4/3piR2^3 (R, R1 and R2 are variables)
    R^3 = R1^3 + R2^3

    S < S1 + S2

    a = 4piR^2 (surface area formula, substitute to have both equations for R)

    4piR^2 < 4piR1^2 + 4piR2^2
    R^2 < R1^2 + R2^2

    Is this correct so far, and now I am stuck with how to put the R^3 in for the R^2 in the second equation.

    Thank you.
  20. Mar 7, 2016 #19
  21. Mar 7, 2016 #20
    Got it. I think I'm having trouble with the distributive properties of exponents with the plus sign its not distributive. So when I plug that in, how do I distribute the exponent here?

    R^3 = R1^3 + R2^3
    convert to R^2
    (R^3)2/3 = R^2
    R^2 = (R1^3 + R2^3)^2/3 ? is this right

    R^2 < R1^2 + R2^2
    (R1^3 + R2^3)^2/3 < R1^2 + R2^2
    0 < R1^2 + R2^2 - (R1^3 + R2^3)^2/3 ? need to simplify
  22. Mar 7, 2016 #21
    If A<B, then A3<B3. So cube both sides of the inequality, and show us what you get.

  23. Mar 7, 2016 #22
    (R1^3 + R2^3)^2/3 < R1^2 + R2^2

    ((R13 + R23)2/3)3 < (R12 + R22)3

    0 < (R12 + R22)3 - ((R13 + R23)2/3)3

    0 < (R12 + R22) - ((R13 + R23)2/3) factor out 3's?
  24. Mar 7, 2016 #23
    With regard to the left-hand side of the inequality ##((x)^a)^b=(x)^{ab}##
  25. Mar 7, 2016 #24
    ((R13 + R23)2/3)3 < (R12 + R22)3

    ((R13 + R23)6/3 < (R12 + R22)3

    (R13 + R23)2 < (R12 + R22)3

    0 < (R12 + R22)3 - (R13 + R23)2
  26. Mar 7, 2016 #25
    Good. Now expand the two terms out.
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